find the greatest value and the least value of the function e^x/x+2

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$$\begin{gathered} \mathrm{Question} \mathrm{must} \mathrm{be} \mathrm{asking} \mathrm{about} \mathrm{local} \mathrm{maxima}/\mathrm{minima} \mathrm{because} \mathrm{golbal} \mathrm{maxima} \\ \mathrm{will} \mathrm{tend} \mathrm{towards} \infty \mathrm{and} \mathrm{global} \mathrm{minima} \mathrm{will} \mathrm{be} \mathrm{tend} \mathrm{towards} -\infty \\ \mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{e}}^{\mathrm{x}}}{\mathrm{x}+2} \\ \mathrm{Differentiate} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\left(\mathrm{x}+2\right).{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}}{{\left(\mathrm{x}+2\right)}^2} \\ \Rightarrow \mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\left(\mathrm{x}+2\right).{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}}{{\left(\mathrm{x}+2\right)}^2}=\frac{{\mathrm{e}}^{\mathrm{x}}\left(\mathrm{x}+1\right)}{{\left(\mathrm{x}+2\right)}^2} \\ \mathrm{Now} \mathrm{let} \mathrm{us} \mathrm{observe} \mathrm{sign} \mathrm{change} \mathrm{of} \mathrm{f}\text{'}\left(\mathrm{x}\right) \\ \frac{{\mathrm{e}}^{\mathrm{x}}}{{\left(\mathrm{x}+2\right)}^2} \mathrm{is} \mathrm{always} \mathrm{positive} \mathrm{hence} \mathrm{sign} \mathrm{change} \mathrm{is} \mathrm{decided} \mathrm{by} \left(\mathrm{x}+1\right) \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)>0 \mathrm{for} \mathrm{x}>-1 \mathrm{and} \mathrm{f}\text{'}\left(\mathrm{x}\right)<0 \mathrm{for} \mathrm{x}<-1 \\ \mathrm{Hence} \mathrm{x}=-1 \mathrm{is} \mathrm{point} \mathrm{of} \mathrm{local} \mathrm{minima}. \\ \mathrm{which} \mathrm{means} \mathrm{local} \mathrm{maxima} \mathrm{does} \mathrm{not} \mathrm{exist}. \\ \mathrm{f}{\left(\mathrm{x}\right)}_{\mathrm{l}.\min }=\mathrm{f}\left(-1\right)=\frac{{\mathrm{e}}^{-1}}{-1+2}=\frac{1}{\mathrm{e}} \end{gathered}$$

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